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Brianna9898

  • one year ago

A floor has two square-shaped designs. The area of the second square-shaped design is nine times greater than the area of the first square-shaped design. Which statement gives the correct relationship between the lengths of the sides of the two squares?

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  1. Brianna9898
    • one year ago
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    these are the options... The length of the side of the second square is 3 times greater than the length of the side of the first square. The length of the side of the second square is 12 times greater than the length of the side of the first square. The length of the side of the second square is 9 times greater than the length of the side of the first square. The length of the side of the second square is 6 times greater than the length of the side of the first square.

  2. Shadowys
    • one year ago
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    since \(A=l^2\) and \(A'=l'^2\) when \(A=9A'\) sub them in to get the relationship between l and l'

  3. Brianna9898
    • one year ago
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    what am i supposed to sub in?

  4. Shadowys
    • one year ago
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    the l and l ' like this: \(A=9A'\) \(l^2 = 9 l'^2\)

  5. Brianna9898
    • one year ago
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    im like really slow and have no idea wut im doing.-_-

  6. Shadowys
    • one year ago
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    lol. i guess i'll try to do it step by step. since \(A=l^2\), \(A_1=l_1^2\) (this is the first square)............(1) \(A_2=l_2^2\) (this is the second square)........(2) since the area of the second square is nine times the first, (it's the only given condition, so we start from that.) \(A_2=9A_1\) subbing (1) and (2) into it, we get, \(l_2^2=9l_1^2\)| what did you get after taking thesquare roots of both sides?

  7. Brianna9898
    • one year ago
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    sorry im back and thank you.

  8. waterineyes
    • one year ago
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    Just take the square root both the sides and tell us what you got as @Shadowys said above..

  9. waterineyes
    • one year ago
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    Getting ?? @Brianna9898

  10. Brianna9898
    • one year ago
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    I have no idea!

  11. waterineyes
    • one year ago
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    Are you getting till here" \[l_2^2=9 l_1^2\]

  12. Brianna9898
    • one year ago
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    why is there a 1 & 2 at the bottom

  13. waterineyes
    • one year ago
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    See when you will take square root you will get like: \[\large \sqrt{l_2^2} = \sqrt{9 l_1^2}\] Can you tell what is this: \[\large \sqrt{l_2^2} = ??\]

  14. Brianna9898
    • one year ago
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    ... and how am I supposed to square root an l ??

  15. waterineyes
    • one year ago
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    Oh that.. \(l_1\) is showing length of first square. \(l_2\) is for length of second square.

  16. waterineyes
    • one year ago
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    you can actually.. See, What is square root of this: \[\large \sqrt{2^2} = ??\]

  17. Brianna9898
    • one year ago
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    2

  18. waterineyes
    • one year ago
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    Similar way what will be square root of this: \[\large \sqrt{l_2^2} = ??\]

  19. Brianna9898
    • one year ago
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    \[l\]

  20. Brianna9898
    • one year ago
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    ??

  21. waterineyes
    • one year ago
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    Or you can say \(l_2\).. Good..

  22. waterineyes
    • one year ago
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    Similarly can you tell for: \[\large \sqrt{9l_1^2} = ??\]

  23. Brianna9898
    • one year ago
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    so you have to keep the 2 at the bottom

  24. Shadowys
    • one year ago
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    the one and two are sub scripts to differentiate between the first length and the second, thus, 1 and 2 respectively.

  25. waterineyes
    • one year ago
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    see, 1 and 2 is differentiating lengths of the two square you are given with, so don't think here of just l think here of \(l_1\) and \(l_2\)..

  26. Brianna9898
    • one year ago
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    so it would be \[9l _{1} ??\]

  27. waterineyes
    • one year ago
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    You forgot to take square root of 9. \(l_1\0 is good though..

  28. waterineyes
    • one year ago
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    What is square root of 9?

  29. Brianna9898
    • one year ago
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    3

  30. waterineyes
    • one year ago
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    Yep after square root you will get like: \[\large l_2 = 3 \times l_1\]

  31. waterineyes
    • one year ago
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    So, which answer choice is this?

  32. Brianna9898
    • one year ago
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    the first one?

  33. waterineyes
    • one year ago
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    Well Done...

  34. waterineyes
    • one year ago
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    And give all the thanks to @Shadowys

  35. Brianna9898
    • one year ago
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    yayyy! thank you guys for helping me!

  36. Brianna9898
    • one year ago
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    I don't even know how to give a medal on this thing

  37. waterineyes
    • one year ago
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    Are you seeing best response after shadows post ?/ Just click that..

  38. Brianna9898
    • one year ago
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    oh okay, can I give a medal to two ppl?

  39. waterineyes
    • one year ago
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    No, just only one..

  40. waterineyes
    • one year ago
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    On one post you can give medal to one only..

  41. Brianna9898
    • one year ago
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    thanx for the medal @Shadowys !

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